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#1 2008-10-21 03:47:02

Danbee
Member
Registered: 2008-08-28
Posts: 21

Solving a Quadratic-Quadratic System by Substitution

(x+5)² +(y+2)²=169
xy-21=0


Please help solve

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#2 2008-10-25 11:35:18

justlookingforthemoment
Moderator
Registered: 2005-05-26
Posts: 2,161

Re: Solving a Quadratic-Quadratic System by Substitution

If you think of this problem visually, you are trying to find the points on a graph where a hyperbola and a circle intersect.

1 -- (x+5)² +(y+2)²=169
2 -- xy - 21=0

xy - 21 = 0
xy = 21
y = 21/x

substitute y = 21/x into equation 1:
(x+5)² +(21/x+2)²=169

and expand:
x² + 10x + 25 + 441/(x²) + 84/x + 4 = 169

which gives:
x^4 + 10x³ - 140x² + 84x + 441 = 0
(x - 7)(x³ + 17x² - 21x - 63) = 0

solve for x:
∴ x ≈ -17.9734 or x ≈ -1.44775 or x ≈ 2.42112 or x = 7

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